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Chapter 3: Atoms: The Building Blocks of Matter CHEMISTRY 1-2 MR. CHUMBLEY Section 1: The Atom: From Philosophical Idea to Scientific Theory The Origin of the Atom Humans have thought about the existence of matter for a very long time In ancient Greece, two competing idea for how matter exists rose to prominence The word “atom” comes from the ancient Greek words, “a-” and “-tomos” This translates to “uncuttables” Democritus’ Particle Theory of Matter Around 400 BC, a Greek philosopher named Democritus is credited for creating the first model of the atom While Democritus did not come up with the idea himself, he made it a fuller idea According to Democritus, “the universe is composed of two elements: the atoms and the void in which they exist and move.” Aristotle’s Continuous Theory of Matter After Democritus’s time, other Greek philosophers came to support a different idea According to Aristotle, matter was continuous and could be broken down indefinitely Aristotle’s idea were more widely accepted as the way in which matter existed, and his opinion was accepted for nearly 2000 years Understanding how Matter Behaves By the late 1700’s, most chemists knew and agreed about elements However, they had difficulty explaining how different substances could combine with one another to form new ones This is the origin of modern chemistry, which focused on quantitative analysis of chemical reactions Law of Conservation of Mass With the help of improved balances and careful measuring, the first law was discovered The law of conservation of mass states that matter is neither created nor destroyed during ordinary chemical reactions or physical changes Law of Definite Proportions The second discovery was that pure substances have a fixed proportion, regardless of how it is made The law of definite proportions states that a chemical compound contains the same elements in exactly the same proportions by mass regardless of the size of the sample or the source of the compound Law of Multiple Proportions Early chemists also knew that the same elements could combine to create different compounds The law of multiple proportions states that if two or more different compounds are composed of the same two elements, then the ratio of the masses of the second element combined with a certain mass of the first element is always a ratio of small whole numbers Dalton’s Atomic Theory In 1808 John Dalton proposed an explanation that included all three laws Dalton’s atomic theory has five points: All matter is composed of extremely small particles called atoms. Atoms of an element are identical in size, mass and other properties; atoms of different elements differ in size, mass, and other properties. Atoms cannot be subdivided, created, or destroyed. Atoms of different elements combine in simple whole-number ratios to form chemical compounds. In chemical reactions, atoms are combined, separated, or rearranged. Democritus to Dalton By using the measurable property of mass, Democritus’s idea was made into Dalton’s scientific theory While we have since learned that not all points of Dalton’s theory hold true, his basic theory has not been discarded, merely modified and improved Section 3: Counting Atoms Atomic Number All atoms of an element have the same number of protons The atomic number of an element is the number of protons of each atom of that element Atomic number is noted by the letter Z In a neutral atom, the number of electrons is equal to the number of protons Isotopes Atoms of the same element can have different number of neutrons Isotopes are atoms of the same element that have different masses Therefore isotopes are atoms with the same number of protons, but different numbers of neutrons Nuclide is a general term for a specific isotope of an element Mass Number Since both protons and neutrons contribute to the mass of the atom, we can find the mass by adding the number of protons to the number of neutrons The mass number is the total number of protons and neutrons that make up the nucleus of an isotope Mass number is noted by the letter A Identifying Isotopes Isotopes can be identified using hyphen notation that combines the name of the element and the mass number For example: uranium-238 carbon-12 hydrogen-3 Identifying Isotopes Isotopes can be identified using the nuclear symbol, which combines the atom’s chemical symbol (X), mass number (A), and atomic number (Z) like so: 𝐴 𝑍𝑋 For example: 238 92𝑈 12 6𝐶 3 1𝐻 Sample Problem A How many protons, electrons, and neutrons are there in an atom of chlorine-37? Step 1: Analyze Given: name and mass number of chlorine-37 Unknown: numbers of protons, electrons, and neutrons Step 2: Plan Atomic number = number of protons = number of electrons Mass number = number of protons + number of neutrons Sample Problem A How many protons, electrons, and neutrons are there in an atom of chlorine-37? Step 3: Solve The mass number of chlorine-37 is 37. Using the periodic table shows the atomic number of chlorine is 17. Atomic number = number of protons = number of electrons = 17 protons and electrons Number of neutrons = mass number – atomic number = 37 – 17 = 20 neutrons An atom of chlorine-37 is made up of 17 electrons, 17 protons, and 20 neutrons Homework Chapter 3, Section 3 Practice (p.76) #1-3 Chapter Review (p.86) #8, 11 Atomic Mass When the mass of a single atom is expressed in kilograms, the value is very small For most calculations it is much easier to use a relative atomic mass For this, an arbitrary measure has been assigned to determine the mass of individual isotopes One unified atomic mass unit, or 1 u, is exactly 1/12 the mass of a carbon-12 atom Atomic Mass However, the mass of isotopes is not simply the mass number expressed in atomic mass units Careful calculation of the masses of different isotopes has shown that differ slightly from the mass number The difference is due to a few factors Protons and neutrons deviate slightly from 1 u The mass of electrons is included in the atomic mass A small amount of mass is changed into energy when proton and neutrons combine to form the nucleus Average Atomic Mass Since most elements occur naturally as isotopes, the mass of that element is determined by the weighted average of those isotopes Average atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes of an element The weighting of the mass values is determined by the percentage that isotope has in natural abundance Average Atomic Mass For example: copper (Cu) is found in two naturally occurring isotopes: copper-63 and copper-65 The mass number, percentage natural abundance, and atomic mass in AMU is shown to the right The average atomic mass is then found by adding the values when the mass of each isotope is multiplied by its natural abundance Isotope Mass # Abundance Atomic Mass (u) Copper-63 63 69.15% 62.929601 Copper-65 65 30.85% 64.927794 Average atomic mass calculation: + 0.6915 × 62.929 601 u 0.3085 × 64.927 794 u 63.55 u Homework Use Figure 3.5 on p. 78 Calculate the average atomic mass of carbon, oxygen, and uranium The Mole Establishing a relative atomic mass makes it possible to know how many atoms of an element are present in a sample of that element with a known mass A mole is the amount of a substance that contains as many particles as there are atoms in exactly 12g of carbon-12 The mole is abbreviated as mol The mole is the base unit of amount in SI Avogadro’s Number Avogadro’s number is the number of particles in exactly one mole of a pure substance This number has been experimentally determined to be 6.022 141 79 × 1023 We will define Avogadro’s number as 6.022 × 1023 This number is named after 19th century scientist, Amedeo Avogadro, who did not actually determine this number Molar Mass Since one mole of a substance contains particles, we can calculate the mass of one mole of that substance The molar mass is the mass of one mole of a pure substance This is a constant value and a property of the material Molar mass is measured in g/mol The value of the molar mass is equivalent in grams to the average atomic mass in atomic mass units Mass and Amount of an Element Mass of an element in grams = × 𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 × 𝟏 𝐦𝐨𝐥 𝟏 𝐦𝐨𝐥 = 𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 Amount of an element in moles = × 𝟏 𝐦𝐨𝐥 × 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬 = 𝟏 𝐦𝐨𝐥 Number of atoms of an element Sample Problem B (p. 80) What is the mass in grams of 3.50 mol of the element copper, Cu? Step 1: Analyze Given: 3.50 mol of Cu Unknown: mass of Cu in grams Step 2: Plan Amount of Cu in moles mass of Cu in grams The mass of an element in grams can be found by multiplying the amount of moles by the molar mass grams Cu moles Cu × = grams Cu moles Cu Sample Problem B (p. 80) What is the mass in grams of 3.50 mol of the element copper, Cu? Step 3: Solve The molar mass of copper from the periodic table is 63.55 g/mol 63.55 g Cu 3.50 mol Cu × = 222 g Cu 1 mol Cu 3.50 mol of copper has a mass of 222 g Sample Problem C (p. 81) A chemist produces 11.9 g of aluminum, Al. How may moles of aluminum were produced? Step 1: Analyze Given: 11.9 g Al Unknown: amount of Al in moles Step 2: Plan mass of Al in grams amount of Al in moles The amount of moles of an element can be found by dividing the mass in grams by the molar mass moles Al grams Al × = moles Al grams Al Sample Problem C (p. 81) A chemist produces 11.9 g of aluminum, Al. How may moles of aluminum were produced? Step 3: Solve The molar mass of aluminum from the periodic table is 26.98 g/mol 1 mol Al 11.9 g Al × = 0.441 mol Al 26.98 g Al 11.9 g of aluminum has 0.441 mols of Al Homework P. 81 Practice Set B: #1-4 Practice Set C: #1-3 Conversions with Avogadro’s Number Similar to how the mass of a substance can be converted to moles, and vice-versa, the number of atoms or particles can be determined using Avogadro’s number Amount of an element in moles = = 𝟏 𝐦𝐨𝐥 × 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬 × 𝟏 𝐦𝐨𝐥 Number of atoms of an element Sample Problem D (p. 82) How many moles of silver, AG, are in 3.01 × 1023 atoms of silver? Step 1: Analyze Given: 3.01 × 1023 atoms of Ag Unknown: amount of Ag in moles Step 2: Plan Number of atoms of Ag amount of Ag in moles The number of atoms is converted to amount in moles by dividing by Avogadro’s number moles Ag Ag atoms × = moles Ag Avogafro′ s number of Ag atoms Sample Problem D (p. 82) How many moles of silver, AG, are in 3.01 × 1023 atoms of silver? Step 3: Solve Avogadro’s number is 6.022 × 1023 . 23 3.01 × 10 1 mol Ag Ag atoms × = 0.500 mol Ag 23 6.022 × 10 Ag atoms 3.01 × 1023 atoms of Ag is 0.500 moles of Ag Sample Problem E (p. 82-83) What is the mass in grams of 1.20 × 1023 atoms of copper, Cu? Step 1: Analyze Given: 1.20 × 1023 atoms of Cu Unknown: mass of Cu in grams Step 2: Plan Number of atoms of Cu amount of Cu in moles mass of Cu in grams The number of atoms is converted to amount in moles by dividing by Avogadro’s number. The amount in moles is then multiplied by molar mass to find the mass in grams Cu atoms × moles Cu grams Cu × = grams Cu Avogadro′ s number of Cu atoms moles Cu Sample Problem E (p. 82-83) What is the mass in grams of 1.20 × 1023 atoms of copper, Cu? Step 3: Solve Avogadro’s number is 6.022 × 1023 . The molar mass of Cu is 63.55 g/mol. 1.20 × 1023 Cu 1 mol Cu 63.55 Cu atoms × × = 1.27 × 10−14 g Cu 23 6.022 × 10 Cu atoms 1 mol Cu 1.20 × 1023 atoms of Cu has a mass of 1.27 × 10-14 grams Homework P. 82 Practice Set D: #1-3 P. 83 Practice Set E: #1-3 Conversions with Avogadro’s Number Similar to how the mass of a substance can be converted to moles, and vice-versa, the number of atoms or particles can be determined using Avogadro’s number Amount of an element in moles = = 𝟏 𝐦𝐨𝐥 × 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬 × 𝟏 𝐦𝐨𝐥 Number of atoms of an element Sample Problem D (p. 82) How many moles of silver, AG, are in 3.01 × 1023 atoms of silver? Step 1: Analyze Given: 3.01 × 1023 atoms of Ag Unknown: amount of Ag in moles Step 2: Plan Number of atoms of Ag amount of Ag in moles The number of atoms is converted to amount in moles by dividing by Avogadro’s number moles Ag Ag atoms × = moles Ag Avogafro′ s number of Ag atoms Sample Problem D (p. 82) How many moles of silver, AG, are in 3.01 × 1023 atoms of silver? Step 3: Solve Avogadro’s number is 6.022 × 1023 . 23 3.01 × 10 1 mol Ag Ag atoms × = 0.500 mol Ag 23 6.022 × 10 Ag atoms 3.01 × 1023 atoms of Ag is 0.500 moles of Ag Sample Problem E (p. 82-83) What is the mass in grams of 1.20 × 1023 atoms of copper, Cu? Step 1: Analyze Given: 1.20 × 1023 atoms of Cu Unknown: mass of Cu in grams Step 2: Plan Number of atoms of Cu amount of Cu in moles mass of Cu in grams The number of atoms is converted to amount in moles by dividing by Avogadro’s number. The amount in moles is then multiplied by molar mass to find the mass in grams Cu atoms × moles Cu grams Cu × = grams Cu Avogadro′ s number of Cu atoms moles Cu Sample Problem E (p. 82-83) What is the mass in grams of 1.20 × 1023 atoms of copper, Cu? Step 3: Solve Avogadro’s number is 6.022 × 1023 . The molar mass of Cu is 63.55 g/mol. 1.20 × 1023 Cu 1 mol Cu 63.55 Cu atoms × × = 1.27 × 10−14 g Cu 23 6.022 × 10 Cu atoms 1 mol Cu 1.20 × 1023 atoms of Cu has a mass of 1.27 × 10-14 grams Homework P. 82 Practice Set D: #1-3 P. 83 Practice Set E: #1-3